Surface Areas and Volumes Class 10 Notes

The concept of surface area and volume for Class 10 is provided here. In this article, we are going to discuss the surface area and volume for different solid shapes such as the cube, cuboid, cone, cylinder, and so on. The surface area can be generally classified into Lateral Surface Area (LSA), Total Surface Area (TSA), and Curved Surface Area (CSA). Here, let us discuss the surface area formulas and volume formulas for different three-dimensional shapes in detail. In this chapter, the combination of different solid shapes can be studied. Also, the procedure to find the volume and its surface area in detail.

To get the solutions for class 10 Maths surface areas and volumes, click on the below link.

• NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes.

Video Lesson on Surface Areas and Volumes Class 10

 

Surface Area and Volume of Cuboid

A cuboid is a region covered by its six rectangular faces. The surface area of a cuboid is equal to the sum of the areas of its six rectangular faces.

Surface Area of the Cuboid

Consider a cuboid whose dimensions are $l\times b\times \mathrm{h,}$ respectively.

The total surface area of the cuboid (TSA) = Sum of the areas of all its six faces

TSA (cuboid) = $2(l\times b)+2(b\times h)+2(l\times h)=2(lb+bh+lh)$

Lateral surface area (LSA) is the area of all the sides apart from the top and bottom faces.

The lateral surface area of the cuboid = Area of face AEHD + Area of face BFGC + Area of face ABFE + Area of face DHGC

LSA (cuboid) = $2(b\times h)+2(l\times h)=2h(l+b)$

Length of diagonal of a cuboid =√(l${}^{}$+ b${}^{}$+ h${}^{}$)

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Volume of a Cuboid

The volume of a cuboid is the space occupied within its six rectangular faces.

Volume of a cuboid $=\left(basearea\right)\times height=\left(lb\right)h=lbh$

Surface Area and Volume of Cube

A cube is a three-dimensional solid that has six square faces, twelve edges and eight vertices.

Surface Area of Cube

As we know, one of the important properties of a cube is length = breadth = height.

If we assume that the length of the cube is “l”, and hence we get

l = breadth = height

So, obviously, here we get,

Breadth = l

Height = l

The total surface area of the cube (TSA) = Sum of the areas of all its six faces.

In case of all faces has an equal area, TSA  of Cube = 6 × area of Square = 6l² square units.

Similarly, the Lateral surface area of cube $=2(l\times l+l\times l)=4l^2$
Note: Diagonal of a cube $=\surd 3l$

Volume of a Cube

Volume of a cube = $basearea\times height$
Since all dimensions of a cube are identical, volume = $l^3$
Where l is the length of the edge of the cube.

To know more about the Volume of Cube and Cuboid, visit here.

For more information on Cube And Cuboid, watch the below video

To know more about the Surface Area of Cube, visit here.

Surface Area and Volume of Cylinder

A cylinder is a solid shape that has two circular bases connected with each other through a lateral surface. Thus, there are three faces, two circular and one lateral, of a cylinder. Based on these dimensions, we can find the surface area and volume of a cylinder.

Surface Area of Cylinder

Take a cylinder of base radius r and height h units. The curved surface of this cylinder, if opened along the diameter (d = 2r) of the circular base can be transformed into a rectangle of length $2\pi r$ and height h units. Thus,

CSA of a cylinder of base radius r and height $h=2\pi \times r\times h$
TSA  of a cylinder of base radius r and height $h=2\pi \times r\times h$ + area of two circular bases
$=2\pi \times r\times h+2\pi r^2$
$=2\pi r(h+r)$

Volume of a Cylinder

Volume of a cylinder = Base area $\times$height = $\left(\pi r^2\right)\times h=\pi r^{2}h$

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To know more about the Surface Area of a Cylinder, visit here.

Surface Area and Volume of Right Circular Cone

A cone is a 3d shape that has one circular base and narrows smoothly from the base to a point called the vertex.

Surface Area of Cone

Consider a right circular cone with slant length l, radius r and height h.

CSA of right circular cone $=\pi rl$
TSA = CSA + area of base $=\pi rl+\pi r^2=\pi r(l+r)$

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Volume of a Right Circular Cone

The volume of a Right circular cone is 1/3 times that of a cylinder of the same height and base.
In other words, 3 cones make a cylinder of the same height and base.
The volume of a Right circular cone $=$$\mathrm{(1/3)\pi }{r}^{2}h$
Where ‘r’ is the radius of the base and ‘h’ is the height of the cone.

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Surface Area and Volume of Sphere

A sphere is a solid that is round in shape, and the points on its surface are equidistant from the center.

Surface Area of Sphere

For a sphere of radius r

Curved Surface Area (CSA) = Total Surface Area (TSA) = $4\pi r^2$

For more information on Sphere And Hemisphere, watch the below video

To learn more about the Surface Area of a Sphere, visit here.

Volume of Sphere

The volume of a sphere of radius r = (4/3)$\pi r^3$

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Surface Area and Volume of Hemisphere

A hemisphere is a shape that is half of the sphere and has one flat surface. The other side of the hemisphere is shaped as a circular bowl. See the figure below.

Surface Area of Hemisphere

We know that the CSA of a sphere  $=4\pi r^2$.

A hemisphere is half of a sphere.
$\therefore$ CSA of a hemisphere of radius r $=2\pi r^2$
Total Surface Area = curved surface area + area of the base circle
$\Rightarrow$TSA $=3\pi r^2$

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Volume of Hemisphere

The volume (V) of a hemisphere will be half of that of a sphere.
$\therefore$ The volume of the hemisphere of radius r $=$$\mathrm{(2/3)\pi }{r}^3$

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Surface Area and Volume of Combination of Solids

The combination of solids explains the shapes formed when two different solids are combined together. Thus, the surface area and volume for such shapes will vary from the other basic solids.

For more information on the Combination of Solids, watch the below video

Surface Area of Combined Figures

Areas of complex figures can be broken down and analysed as simpler known shapes. By finding the areas of these known shapes, we can find out the required area of the unknown figure.
Example: 2 cubes each of volume 64 $c{m}^3$  are joined end to end. Find the surface area of the resulting cuboid.
Length of each cube $={\mathrm{64(1/3)}}^{}$$=4cm$
Since these cubes are joined adjacently,  they form a cuboid whose length l = 8 cm. But height and breadth will remain the same = 4 cm.

$\therefore$ The new surface area, TSA = 2(lb + bh + lh)

TSA = 2 (8 x 4 + 4 x 4 + 8 x 4)

= 2(32 + 16 + 32)

= 2 (80)

TSA = 160 cm²

To learn more about the surface area of the combination of solids, click here.

Volume of Combined Solids

The volume of complex objects can be simplified by visualising them as a combination of shapes of known solids.
Example: A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 3 cm and the height of the cone is equal to 5 cm.
This can be visualised as follows :

V(solid) = V(Cone) + V(hemisphere)

V(solid) = (1/3)$\pi r^2$h + (2/3)$\pi r^3$

V(solid) = (1/3)$\mathrm{\pi (9)(5)}$+ (2/3)$\mathrm{\pi (27)}$

V(solid) = 33$\mathrm{\pi \; cm3}$

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Video Lesson on Combination of Solids

 

Surface Area and Volume of Frustum of a cone

When a solid is cut by a plane, then another form of solid is formed. One such form of a solid is the frustum of a cone, which is formed when a plane cuts a cone parallelly to the base of the cone. Let us discuss its surface area and volume here.

If a right circular cone is sliced by a plane parallel to its base, then the part with the two circular bases is called a Frustum.

Surface Area of a Frustum

CSA of frustum $=\pi (r_1+r_2)l$,  where $l= \surd [h2+(r2 – r1)2]$

TSA of the frustum is the CSA + the areas of the two circular faces = $\pi (r_1+r_2)l+\pi ({\mathrm{r12}}_{}+{\mathrm{r22}}^{})$

Example:

Find the curved surface area of the frustum if the slant height of the cone’s frustum is 4 cm. Also, given that the circumferences of its circular ends are 6 cm and 18 cm.

Solution:

Given: Slant height, l = 4 cm

The circumferences of the two circular ends are 6 cm and 18 cm.

Let the radius of the two circular ends be r₁ and r₂.

Now, we have to find r₁ and r₂.

Finding r₁:

Circumference of one end = 6 cm

2πr₁ = 6 

r₁ = 6/2π

r₁ = 3/π

Finding r₂:

Circumference of the other end = 18 cm

2πr₂ = 18 

r₂ = 18/2π

r₂ = 9/π

We know that the formula to find the curved surface area of frustum is π(r₁ + r₂)l square units.

Thus by substituting the known values, we get

CSA of frustum = π[(3/π) + (9/π)]4

CSA of frustum = 4π(12/π)

CSA of frustum = 48 cm².

Therefore, the curved surface area of the frustum is 48 cm².

Volume of a Frustum

The volume of a frustum of a cone $=$$\mathrm{(1/3)\pi }h(r12 + r22+r_1{r}_2)$

Example:

A glass is in the shape of the frustum of a cone with a height of 14 cm. The diameters of its two circular ends are 2 cm and 4 cm. Find the volume of the glass.

Solution:

Given: Height of the frustum = 14 cm

The diameters of the two circular ends are 2 cm and 4 cm.

So, the radius will be 1 cm and 2 cm.

Let r₁ = 1 cm and r₂ = 2 cm

Since the glass is in the shape of a frustum, the volume of a glass = volume of a frustum.

Therefore, the volume of glass = (⅓)πh (r₁² + r₂² + r₁r₂) cubic units.

Now, substitute the values in the formula, we get

Volume of a glass, V = (⅓) π(14)[1² + 2² + (1)(2)]

V = (⅓) (22/7)(14) [1 + 4 + 2]

V = (⅓) (22/7)(14)(7)

V = [22 × 14 × 7] / [3 × 7]

V = (22 × 14 ) / 3

V = 308 / 3

V = 102. 67

Hence, the volume of a glass is 102. 67 cm³.

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Conversion of Solid from One Shape to Another

When a solid is converted into another solid of a different shape (by melting or casting), the volume remains constant.

Example:

A metallic sphere with a radius of 4.2 cm is melted and recast in the shape of a cylinder with a radius of 6 cm. Determine the cylinder’s height.

Solution:

Given that, the sphere is melted and recast into a cylinder.

So, we can write, 

Sphere’s volume = Cylinder’s volume. 

As we know, the volume of a sphere = (4/3)πr³ cubic units.

And the volume of a cylinder = πr²h cubic units.

So, we can write:

(4/3)πr³ =  πr²h …(1)

Given that, the radius of the metallic sphere = 4.2 cm.

Radius of the cylinder = 6 cm.

Now, substitute the values in (1), we get

(4/3)π(4.2)³ =  π(6)²h 

(4/3)(4.2)³ = 36h

Hence, h = [(4/3)(4.2)³]/36

h = [4 × 4.2 × 4.2 × 4.2 ]/[3 × 36]

By simplifying the above expression, we get

h = 2.744 cm

Hence, the height of the cylinder is 2.744 cm.